Explained: "Solve the integral by substitution. ∫(z2+2z+1)43dz"

CheemnCatelvew

Answered question

2021-11-05

Solve the integral by substitution.

\int {(z^{2} + 2 z + 1)}^{\frac{4}{3}} d z

Fatema Sutton

Skilled2021-11-06Added 88 answers

Step 1
The given integral is,

I = \int {(z^{2} + 2 z + 1)}^{\frac{4}{3}} d z

On simplifying the term under integral, we get

z^{2} + 2 z + 1 = {(z)}^{2} + 2 \cdot z \cdot 1 + {(1)}^{2}

= {(z + 1)}^{2} \dots . [a^{2} + 2 a b + b^{2} = {(a + b)}^{2}]

Hence, the integral becomes,

I = \int {({(z + 1)}^{2})}^{\frac{4}{3}} d z

= \int {(z + 1)}^{\frac{8}{3}} d z

Step 2
On substituting z+1 as t
z+1=t
dz=dt
Hence, the given integral becomes,

I = \int t^{\frac{8}{3}} d t

Using the power rule of integration, we get

I = \int t^{\frac{8}{3}} d t

= [\frac{t^{\frac{8}{3} + 1}}{\frac{8}{3} + 1}] + C

= [\frac{t^{\frac{11}{3}}}{(\frac{11}{3})}] + C

= \frac{3 t^{\frac{11}{3}}}{11} + C

Step 3
Putting back the value of t, we get

\frac{3 t^{\frac{11}{3}}}{11} + C = \frac{3 {(z + 1)}^{\frac{11}{3}}}{11} + C

Therefore, the expression for the given integral is

\frac{3 {(z + 1)}^{\frac{11}{3}}}{11} + C

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